Calculating Displacement in Two-Dimensional Motion

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The discussion centers on calculating displacement in two-dimensional motion using vector components. The problem involves traveling at a speed of 1.2 m/sec for 45 minutes at a heading of 45 degrees, then changing direction to 330 degrees for 32 minutes. The user initially calculates the displacement for the first vector and arrives at 3240 meters, questioning its validity. Responses clarify that this value is reasonable given the time and speed, confirming that the calculation is correct. Overall, the conversation emphasizes understanding vector addition in two-dimensional motion.
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I'm having trouble with adding two-dimensional motion. Here's an example of the type of problem I'm having trouble with:

A map gives you the following directions:

Travel 45 mintues at a speed of 1.2 m/sec and a heading of 45 degrees. Turn to a heading of 330 degrees and travel at the same speed for 32 minutes.
What will your displacement be if you follow these directions?

I was taught to use this equation:
\Deltax = initial velocity * time + 1/2 * acceleration * time^2
and
Ax= initial velocity * cos()
Ay= initial velocity * sin()
and the same for the the second vector.

I started out with trying to find the first vector or vector A as I called it.
\Deltax= 1.2 m/sec(2700 sec) + 1/2 (0)(2700 sec)^2
and got \Deltax=3240m which I thought was a bit high, so I don't know if I'm doing something wrong or if it should really be that high.
 
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Welcome to PF!

Hi Gamegoofs! Welcome to PF! :smile:

(have a delta: ∆ :wink:)
Gamegoofs2 said:
Travel 45 mintues at a speed of 1.2 m/sec and a heading of 45 degrees.

I started out with trying to find the first vector or vector A as I called it.
\Deltax= 1.2 m/sec(2700 sec) + 1/2 (0)(2700 sec)^2
and got \Deltax=3240m which I thought was a bit high, so I don't know if I'm doing something wrong or if it should really be that high.

No, that's ok …

and it's not high, it's only 3km after nearly an hour! :wink:
 
Thanks so much!
 
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